1.

In the given figure, ∆ODC~∆OBA, ∠BOC = 115° and ∠CDO = 70° . Find(i) ∠DCO (ii) ∠DCO (iii) ∠OAB (iv) ∠OBA.

Answer»

(i) It is given that DB is a straight line. 

Therefore, 

∠DOC + ∠COB = 180°

∠DOC = 180° − 115° = 65°

(ii) In ∆ DOC, we have: 

∠ODC + ∠DCO + ∠DOC = 180°

Therefore, 

70° + ∠DCO + 65° = 180°

⇒ ∠DCO = 180 − 70 − 65 = 45°

(iii) It is given that ∆ ODC - ∆ OBA 

Therefore, ∠OAB = ∠OCD = 45°

(iv) Again, ∆ ODC- ∆ OBA 

Therefore, 

∠OBA = ∠ODC = 70°



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